Question

A circular disc A of radius 18 cm is made from an iron plate of thickness 2 cm and another circular disc B of radius 72 cm is made from an iron plate of thickness 0.5 cm. The relation between the moments of inertia IA and IB is.........

guys plz ans this..​

Answer 1

Answer:

The relation between $I_B$ and $I_A$ :

$I_B=64 \times I_A$

Explanation:

Given two circular discs:

Disc A has following dimensions:

Radius, $r_A = 18 cm$

Thickness, $h_A = 2 cm$

Let the density of iron = $\rho\ kg/cm^3$

Mass of disc A, $m_A=Density \times Volume$

A circular iron plate is in the form of a cylinder.

So,

$m_A=\rho \times \pi r_A^2\times h_A\\\Rightarrow m_A=\rho \times \pi 18^2\times 2\ kg$

Now, Intertia is given by the formula:

$I =\dfrac{1}{2} mr^2$

$I_A =\dfrac{1}{2} m_Ar_A^2\\\Rightarrow I_A =\dfrac{1}{2} \rho \times \pi \times 18^2 \times 2\times 18^2\\\Rightarrow I_A =2 \times \rho \pi \times 18^4 \times...... (1)$

Disc B has following dimensions:

Radius, $r_B = 72 cm$

Thickness, $h_B = 0.5 cm$

It is also made up of iron.

Let the density of iron = $\rho\ kg/cm^3$

Mass of disc B, $m_B=Density \times Volume$

A circular iron plate is in the form of a cylinder.

So,

$m_B=\rho \times \pi r_B^2\times h_B\\\Rightarrow m_B=\rho \times \pi \times 72^2\times 0.5\ kg$

Now, Intertia is given by the formula:

$I =\dfrac{1}{2} mr^2$

$I_B =\dfrac{1}{2} m_Br_B^2\\\Rightarrow I_B =\dfrac{1}{2} \rho \times \pi \times 72^2 \times 0.5\times 72^2\\\Rightarrow I_B =\dfrac{1}4\rho \pi \times 72^3 \times 18 \times 4 \\\Rightarrow I_B =\rho \pi \times (18 \times 4)^3 \times 18 \\\Rightarrow I_B =64 \times \rho \pi \times 18^4...... (2)$

By (1) and (2):

$I_B=64 \times I_A$

So, The relation between $I_B$ and $I_A$ :

$I_B=64 \times I_A$